11.3 Graphing with Intercepts

Learning objectives

Learning objectives

• Identify the intercepts on a graph

• Find the intercepts from an equation of a line

• Graph a line using the intercepts

• Choose the most convenient method to graph a line

Identify the Intercepts on a Graph

Every linear equation has a unique line that represents all the solutions of the equation. When graphing a line by plotting points, each person who graphs the line can choose any three points, so two people graphing the line might use different sets of points.

At first glance, their two lines might appear different since they would have different points labeled. But if all the work was done correctly, the lines will be exactly the same line. One way to recognize that they are indeed the same line is to focus on where the line crosses the axes. Each of these points is called an intercept of the line.

Let’s look at the graph of the lines shown in Figure 11.14.

First, notice where each of these lines crosses the x- axis:

Do you see a pattern?

For each row, the y- coordinate of the point where the line crosses the x- axis is zero. The point where the line crosses the x- axis has the form $(a,0)$; and is called the x-intercept of the line. The x- intercept occurs when y is zero.

Now, let's look at the points where these lines cross the y-axis.

Example 11.23

Solution

1. Use the intercept definitions. The $x$-intercept has $y=0$, while the $y$-intercept has $x=0$, so read each axis crossing as an ordered pair.
2. Read the first line. It crosses the $x$-axis at $(4,0)$ and the $y$-axis at $(0,2)$, giving those two intercepts.
3. Read the second line. It crosses the $x$-axis at $(2,0)$ and the $y$-axis at $(0,-6)$, so the negative value is part of the $y$-intercept.
4. Read the third line. It crosses the $x$-axis at $(-5,0)$ and the $y$-axis at $(0,-5)$, so both intercepts of this line lie on the negative sides.
5. Check the zero coordinates. Every reported $x$-intercept has second coordinate $0$, and every reported $y$-intercept has first coordinate $0$, matching the axes crossed.

Takeaway: A line crosses the $x$-axis where $y=0$ and the $y$-axis where $x=0$, so reading a graph gives its intercepts directly: $(4,0)$ and $(0,2)$ for the first line shown.

Find the Intercepts from an Equation of a Line

Recognizing that the $x\mathrm{-intercept}$ occurs when $y$ is zero and that the $y\mathrm{-intercept}$ occurs when $x$ is zero gives us a method to find the intercepts of a line from its equation. To find the $x\mathrm{-intercept,}$ let $y=0$ and solve for $x.$ To find the $y\mathrm{-intercept},$ let $x=0$ and solve for $y.$

Example 11.24

Solution

1. Find the $x$-intercept. Set $y=0$ in $2x+y=6$, giving $2x+0=6$, so $2x=6$ and $x=3$.
2. Find the $y$-intercept. Set $x=0$, giving $2(0)+y=6$, so $0+y=6$ and $y=6$.
3. Record the intercept points. The $x$-intercept is $(3,0)$ and the $y$-intercept is $(0,6)$.
4. Check the displayed work. The figures show these substitutions, the simplified equations $2x=6$ and $y=6$, and the summary table listing the ordered pairs $(3,0)$ and $(0,6)$.

Takeaway: Set the coordinate that should be zero on the desired axis, then solve the resulting one-variable equation.

Example 11.25

Solution

1. Set $y$ equal to zero. Substitution into $4x-3y=12$ gives $4x-3(0)=12$, so $4x=12$ and $x=3$.
2. Record the $x$-intercept. Because the point lies on the $x$-axis, its ordered pair is $(3,0)$.
3. Set $x$ equal to zero. Substitution gives $4(0)-3y=12$, so $-3y=12$ and dividing by $-3$ gives $y=-4$.
4. Record the $y$-intercept. Because the point lies on the $y$-axis, its ordered pair is $(0,-4)$.
5. Verify both coordinates. The pairs $(3,0)$ and $(0,-4)$ make the original equation equal $12$, confirming the two intercepts.

Takeaway: For $4x-3y=12$, substituting $y=0$ gives the $x$-intercept $(3,0)$ and substituting $x=0$ gives the $y$-intercept $(0,-4)$ — two one-variable equations locate the whole line.

Graph a Line Using the Intercepts

To graph a linear equation by plotting points, you can use the intercepts as two of your three points. Find the two intercepts, and then a third point to ensure accuracy, and draw the line. This method is often the quickest way to graph a line.

Example 11.26

Solution

1. Find the $x$-intercept. Set $y=0$ in $-x+2y=6$, giving $-x=6$ and $x=-6$, so the intercept is $(-6,0)$.
2. Find the $y$-intercept. Set $x=0$, giving $2y=6$ and $y=3$, so the intercept is $(0,3)$.
3. Find a third point. Choose $x=2$, then $-2+2y=6$ gives $2y=8$ and $y=4$, producing $(2,4)$.
4. Plot the three points. Mark $(-6,0)$, $(0,3)$, and $(2,4)$ on the coordinate plane.
5. Check and draw. The three points lie on one straight line, so draw the line through them to graph $-x+2y=6$.

Takeaway: Setting $y=0$ and then $x=0$ in $-x+2y=6$ yields $(-6,0)$ and $(0,3)$, and a third point such as $(2,4)$ confirms the line before graphing it.

Example 11.27

Solution

1. Find the $x$-intercept. Set $y=0$ in $4x-3y=12$, giving $4x=12$ and $x=3$, so the intercept is $(3,0)$.
2. Find the $y$-intercept. Set $x=0$, giving $-3y=12$ and $y=-4$, so the intercept is $(0,-4)$.
3. Find a third point. Choose $x=6$, then $4(6)-3y=12$ gives $24-3y=12$, so $y=4$ and the point is $(6,4)$.
4. Plot the three points. Mark $(3,0)$, $(0,-4)$, and $(6,4)$ on the coordinate plane.
5. Check and draw. These points lie on one straight line, so connect them to complete the graph of $4x-3y=12$.

Takeaway: Two intercepts plus one check point — $(3,0)$, $(0,-4)$ and $(6,4)$ for $4x-3y=12$ — are enough to draw a straight line accurately.

Example 11.28

Solution

1. Identify the common intercept. Substituting $x=0$ into $y=5x$ gives $y=0$, so both intercepts coincide at the origin $(0,0)$.
2. Choose another $x$-value. Let $x=1$, which gives $y=5(1)=5$ and the additional point $(1,5)$.
3. Choose a negative $x$-value. Let $x=-1$, which gives $y=5(-1)=-5$ and the additional point $(-1,-5)$.
4. Plot and connect the points. Plot $(0,0)$, $(1,5)$, and $(-1,-5)$; because they lie on one straight line, draw the line through them.

Takeaway: Because $y=5x$ passes through the origin, both intercepts collapse to $(0,0)$, so extra points such as $(1,5)$ and $(-1,-5)$ are required to draw the line.

Choose the Most Convenient Method to Graph a Line

While we could graph any linear equation by plotting points, it may not always be the most convenient method. This table shows six of equations we’ve graphed in this chapter, and the methods we used to graph them.

What is it about the form of equation that can help us choose the most convenient method to graph its line?

Notice that in equations $1 and $2, y is isolated on one side of the equation, and its coefficient is 1. We found points by substituting values for x on the right side of the equation and then simplifying to get the corresponding y- values.

Equations $3 and $4 each have just one variable. Remember, in this kind of equation the value of that one variable is constant; it does not depend on the value of the other variable. Equations of this form have graphs that are vertical or horizontal lines.

In equations $5 and $6, both x and y are on the same side of the equation. These two equations are of the form $Ax+By=C$. We substituted $y=0$ and $x=0$ to find the x- and y- intercepts, and then found a third point by choosing a value for x or y.

This leads to the following strategy for choosing the most convenient method to graph a line.

Example 11.29

Solution

1. Classify the first equation. Since $y=-3$ contains only $y$, graph it as a horizontal line crossing the $y$-axis at $-3$.
2. Classify the second equation. The form $4x-6y=12$ contains both variables, so find the two intercepts and one additional point.
3. Classify the third equation. Since $x=2$ contains only $x$, graph it as a vertical line crossing the $x$-axis at $2$.
4. Classify the fourth equation. Because $y$ is isolated in $y=\frac{2}{5}x-1$, choose three convenient $x$-values and plot the resulting points.
5. Compare the methods. Each method uses the structure of the equation to minimize the calculations needed to locate the line accurately.

Takeaway: Match the equation's form to the quickest method: a lone variable gives a horizontal or vertical line, $Ax+By=C$ invites the intercepts, and an isolated $y$ invites three plotted points.

Section 11.3 Exercises

Section 11.3 Exercises

Identify the Intercepts on a Graph

In the following exercises, find the $x-$ and $y-$ intercepts.

Find the $x$ and $y$ Intercepts from an Equation of a Line

In the following exercises, find the intercepts.

Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

Choose the Most Convenient Method to Graph a Line

In the following exercises, identify the most convenient method to graph each line.

Everyday Math

Writing Exercises

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

1. Find the $x-$ and $y\mathrm{-intercepts}$ of the graph: $x-y=2.$

2. Find the $x-$ and $y\mathrm{-intercepts}$ of the graph: $2x+3y=6.$

3. Find the intercepts: $3x+y=12$

1. Find the intercepts: $x+4y=8$

2. Find the intercepts of the line: $3x-4y=12.$

3. Find the intercepts of the line: $2x-4y=8.$

1. Identify the most convenient method to graph each line:

1. ⓐ $\,3x+2y=12$

2. ⓑ $\,y=4$

3. ⓒ $\,y=\frac{1}{5}x-4$

4. ⓓ $\,x=-7$

2. Identify the most convenient method to graph each line:

1. ⓐ $\,x=6$

2. ⓑ $\,y=-\frac{3}{4}x+1$

3. ⓒ $\,y=-8$

4. ⓓ $\,4x-3y=-1$

3. $x+y=4$

1. $x+y=-2$

2. $x-y=5$

3. $x-y=-3$

1. $x+2y=8$

2. $3x+y=6$

3. $x-3y=12$

1. $4x-y=8$

2. $2x+5y=10$

3. $3x-2y=12$

1. $y=\frac{1}{3}x-1$

2. $y=\frac{1}{5}x+2$

3. $y=3x$

1. $y=-4x$

2. $2x+4y=12$

3. $x=2$

1. $y=5$

2. $y=-3x+4$

3. $x-y=5$

1. $y=\frac{2}{3}x-1$

2. $y=-3$

3. $3x-2y=-12$

1. $y=-\frac{1}{4}x+3$

2. Road trip Damien is driving from Chicago to Denver, a distance of $1,000$ miles. The $x\mathrm{-axis}$ on the graph below shows the time in hours since Damien left Chicago. The $y\mathrm{-axis}$ represents the distance he has left to drive.

ⓐ Find the $x-$ and $y-$ intercepts

ⓑ Explain what the $x-$ and $y-$ intercepts mean for Damien.

3. How do you find the $x\mathrm{-intercept}$ of the graph of $3x-2y=6?$

1. Do you prefer to graph the equation $4x+y=-4$ by plotting points or intercepts? Why?